CHE 163 HW CH#12
04, 10, 12, 14, 16, 24, 28, 32, 34, 38, 42, 46, 50, 54, 56, 58, 60, 86, 90, 94, 96, 112, 114.
12-04
Because 4 A and 4 B ions occupy the corners, and because the portion of the unit cell contains 18 of
each corner ion, the numbers of A and B cations in the unit cell are
4 A cations  18 = 12 A cation in the unit cell
4 B cations  18 = 12 B cation in the unit cell
Because 4 X ions occupy the edges, and because the unit cell contains 14 of each edge ion, the number
of X anions in the unit cell is
4 X anions  14 = 1 X anion in the unit cell
This gives a total of one anion and one cation per unit cell.
12-10
The number of small spheres in the unit cell is (8  18 )  (6  12 )  4. The number of large spheres in
the unit cell is (8  1) = 8.
12-12
The number of sulfur atoms in the unit cell is
(8  18 )  (6  12 )  4
The number of zinc atoms in the unit cell is
8 tetrahedral  12  4
The formula for the compound is ZnS.
12-14
Along the body diagonal we have 2rCl–  2rCs where r is the radius of the respective ions. Setting
this equal to the length of the body diagonal and substituting in (the unit cell edge length) and the
radius of the Cl– ion gives
2rCl–  2rCs  3
 2 181 pm   2rCs

  412 pm  3
2rCs  351.6 pm
rCs  176 pm
12-16
133
 0.782
170
72
 0.54
For MgF2, the radius ratio is
133
100
 0.752
For CaF2, the radius ratio is
133
118
 0.887
For SrF2, the radius ratio is
133
135
 1.02
For BaF2, the radius ratio is
133
The fluorite structure is not possible for any of these salts as predicted by the radius ratio rule.
However, both CaF2 and SrF2 adopt the fluorite structure. The elements Cs (red), Mg (orange), and
Ba (purple) do not.
For CsF, the radius ratio is
12-24
Under high pressure and low temperature, the filled 1s orbitals on many He atoms would form a
filled valence band. The next orbitals in energy, the 2s orbitals, could form the conduction band. If
the conditions are such that the valence band and the conduction band overlap (as shown in Figure
12.3 for zinc), then solid helium could conduct electricity.
12-28
Because germanium holds its electrons less tightly as indicated by its lower ionization energy, its
presence in silicon might enhance the electrical conductivity of silicon.
12-32
(a) Because boron has one fewer valence electron than carbon, the B-doped diamond is an example of a
p-type semiconductor.
6.626  10–34 J  s  3.00  108 m/s
 2.94  10–19 J
6.75  10–7 m
2.94  10–19 J 6.022  1023 photons 1 kJ


 177 kJ/mol
photon
mol
1000 J
(c) Eg 
12-34
6.626  10 –34 J  s  3.00  108 m/s
 4.96  10 –7 m or 496 nm
 241.1 kJ 1000 J

1 mol
 mol  kJ  6.022  1023 photons 
For AlP  
For GaP  
For InP  
6.626  10 –34 J  s  3.00  108 m/s
 5.54  10 –7 m or 554 nm
 216.0 kJ 1000 J

1 mol
 mol  kJ  6.022  1023 photons 
6.626  10 –34 J  s  3.00  108 m/s
 9.77  10 –7 m or 977 nm
 122.5 kJ 1000 J

1 mol
 mol  kJ  6.022  1023 photons 
12-38
From Figure 12.16 we can see that the packing in the bcc unit cell is less dense (more open spaces)
than in the fcc unit cell. Therefore, the fcc unit cell has greater packing efficiency.
12-42
In the simple cubic structure, the edge length
Pythagorean theorem:
Face diagonal 
The body diagonal can then be determined:
Body diagonal 
 2r 2   2r
is 2r and the face diagonal can be calculated with the
2
2


2
2
 2
2

2  2r 2
 4r 2  8r 2  r 12  2r 3 or 3.46r
In the face-centered cubic structure, the edge length is and the face diagonal is
12.16). The face diagonal is also equal to 4r (r = atomic radius), so
4r
4r  2 or 
2
Substituting into the Pythagorean theorem equation:
2 (from Figure
2
16r 2
 4r 
2
Body diagonal  
  4r  
 16r 2  8r 2  16r 2  r 24  2r 6 or 4.90r

 2
2
12-46

4r
2

4  143 pm
2
 404 pm
12-50
The disruption of the gold lattice by the smaller Ni atoms in white gold does not allow the gold
atoms to slip past each other as easily compared to pure gold. Therefore, pure gold is more malleable.
12-54
The radius ratio of these alloys fall in the range
75 pm
88 pm
 0.42 to
 0.71
180 pm
124 pm
12-56
(a) For 8 A atoms at the corners and 1 B atom in the center of the unit cell:
( 18  8 A atoms) + (1  1 B atom) = AB
(b) If the atom positions were reversed, we would get the same formula for the alloy, AB.
12-58
radius of boron atom
 0.73
radius of closest-packed atom
radius of boron 88 pm
Radius of atom 

 121 pm
0.73
0.73
12-60
For Sn3Hg
rSn 140 pm

 100  92.7% or a 7.3% difference
rHg 151 pm
For bronze
rCu 128 pm

 100  91.4% or an 8.6% difference
rSn 140 pm
Sn3Hg has a smaller mismatch in atomic size.
12-86
The number of cubic holes in the simple cubic unit cell is 1. The number of F– anions in the unit cell
is also 1.
12-90
(a) The rock salt ( ) form of MnS has an fcc arrangement of S2– ions with all of the 4 octahedral
holes in the unit cell occupied by Mn2+. The sphalerite ( ) form of MnS also has an fcc arrangement
of S2– ions with half of the 8 tetrahedral holes in the unit cell occupied by Mn 2+.
(b) The radius ratio, being below 0.41, predicts that Mn 2+ fits into the tetrahedral holes in the unit
cell.
12-94

39.098 g  
79.904 g  
1 mol
Mass of the unit cell   4 
 4 

 7.9045  10 –22 g



23




mol
mol
6.022

10


7.9045  10 –22 g
Volume 
 2.8743  10 –22 cm3
3
2.75 g/cm
Edge length  3 2.8743  10 –22 cm 3  6.60  10 –8 cm
1 pm
Edge length in picometers  6.60  10 –8 cm 
 660 pm
1  10 –10 cm
12-96
In ceramic materials like TiO2, the band gap between the valence band and the conduction band is
very large and therefore the substance is an electrical insulator. Titanium, however, being a metal
and having partially filled d orbitals, has essentially no band gap (similar to Cu in Figure 12.2) and
so is an electrical conductor.
12-112
Bromide ions are larger than chloride ions so the distance between AgBr layers is greater than
between AgCl layers. Rearranging the Bragg equation to highlight the relationship between  and d
gives
n
 sin 
2d
From this equation we see that as d (distance between the atom layers) increases, sin  decreases
and so 2 will also decrease. The compound expected to diffract X-rays through larger values of 2
is AgCl.
12-114
To determine n we notice that 32.84˚ is 1.55 times 21.19˚ and that 46.30˚ is 2.18 times 21.19˚. The
pattern here gives n = 2 (21.19˚), n = 3 (32.84˚), and n = 4 (46.30˚).
The spacings between the layers are
n
2  154 pm
d

 426 pm
2sin  2 sin 21.19
3  154 pm

 426 pm
2 sin 32.84
4  154 pm

 426 pm
2 sin 46.30
12.115.
sin  
1  154 pm
 0.04151
2  1855 pm
  2.38
2  4.76